Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

So let $S_N$ be the symmetric group of degree $N$. We think of it as a permutation group via its natural action on the set $T=\{1,2,\ldots,N\}$.

Say that $H\leq S_N$ is a subgroup which acts transitively on $T$. However, I DONT'T WANT to assume necessarily that $H$ is primitive (that is the whole point of my question). Assume furthermore that there is an onto group homomorphism $$ f:H\rightarrow S_n $$ where $n=\lfloor{N/2}\rfloor$. In fact, as was pointed out by Schmidt, the existence of this onto group homomorphism implies that $H$ is imprimitive.

In general, one cannot rule out the existence of such an $H$. For example one could have $H=S_n\ltimes\mathbf{F}_2^n$ when $N$ is even and $n=\frac{N}{2}$. Here, $S_n$ acts in the natural way by permutation on the coordinates of $\mathbf{F}_2^n$. Note that by construction, $H$ acts transitively on $T$ and it admits an onto group homomorphism on $S_n$.

Furthermore, suppose that I can produce " a lot of elements " in $H$ which contain a cycle of length $r$ in their cycle presentations (their writing as a product of disjoint cycles of $T$) for $r>n$. Then may I conclude that such an $H$ does not exist?

Q1: Is there some kind of results that would allow me to conclude that $H\supseteq A_N$, so that this would contradict the imprimitivity and therefore rule out the existence of such an $H$?

For example here is one key result which is good to know: if $H$ is assumed to be primitive and contains a cycle of length $\ell$ with $2\leq \ell\leq N-7$ ($\ell$ not necessarily prime) then combining classical results on permutation group theory one may show that $H\supseteq A_N$. However, since in my setting $H$ is imprimitive I cannot apply this result.

Q2: Do we have a good understanding of the tree of subgroups of $S_N$, especially the maximal subgroups?

Q3: Is there some kind of probabilistic result that could be used in my context?

share|improve this question
Be careful that the subgroup H you are talking about exists. There is no onto group homomorphism from the symmetric group on 2n points to the symmetric group on n points unless n=1 or n=2. In particular, unless n=1 or n=2, H never contains the alternating group of degree N=2n. –  Jack Schmidt Mar 21 '11 at 20:56
Well take $H=S_n\rtimes\mathbf{F}_2^n$ with $N=2n$, this certainly have an onto group homomorphism to $S_n$. –  Hugo Chapdelaine Mar 21 '11 at 22:09
change $\rtimes$ by $\ltimes$ –  Hugo Chapdelaine Mar 21 '11 at 22:10
And such an H never contains the alternating group of degree N. –  Jack Schmidt Mar 22 '11 at 2:42
Looks good. Your solution looks good. The maximal subgroups of symmetric groups are a little bit complicated, but for the most part are well-understood. Let me know if you want an answer about the subgroups of symmetric groups or maximal subgroups of symmetric groups, but I think you've got what you need. –  Jack Schmidt Mar 22 '11 at 23:18

1 Answer 1

up vote 2 down vote accepted

Well I think I have more or less an answer to my question. I have shown that the set of all maximal imprimitive transitive subgroups $H\leq S_N$ is of the form $$ S_{N/r}^{r}\rtimes S_r $$ for $r|N$ and where $S_r$ acts by permutation on the coordinates of $S_{N/r}^r$. So since I have an onto group homomorphism $$ f:H\rightarrow S_n $$ I must conclude that $H\subseteq S_{2}^{n}\rtimes S_n$ and that $H\supseteq S_n$. Finally, since I can produce an element $\tau\in H$ that has a cycle of length larger than $n$ which appears in its cycle presentation I may conclude that $H$ is not contained in any maximal transitive imprimitive subgroups of $S_N$ and therefore by maximality this implies that $H=S_N$. But this is absurd since it contradicts the imprimitivity of $H$. Therefore such an $H$ does not exist.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.